How to evaluate $\int \frac { \sin x+\cos x }{ \sin^4 x+\cos^4x}\, dx$?
How can one find $$\int \frac { \sin x+\cos x }{ \sin^4 x+\cos^4x}\, dx?$$
Solution 1:
HINT:
$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-\dfrac{\sin^22x}2$
Set $\dfrac{d(\cos x+\sin x)}{dx}=-\sin x+\cos x=u\implies u^2=1-\sin2x$
Solution 2:
Split the integral as $$ \int\frac {\cos x}{\sin^4 x + (1 - \sin^2 x)^2} dx + \int \frac{\sin x}{(1 - \cos^2 x)^2 + \cos^4 x} dx $$ Compute the integrals with the substitutions $u = \sin x$ and $u = \cos x$ respectively.